In a triangle ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm.

Question:

In a triangle ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of triangle ABC and hence its altitude on AC.

Solution:

Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.

So given,

a = 15 cm

b = 13 cm

c = 14 cm

By using Heron's Formula

The Area of the triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

Semi perimeter of a triangle = 2s

2s = a + b + c

s = (a + b + c)/2

s = (15 + 13 + 14)/2

s = 21 cm

Therefore,

Area of the triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

$=\sqrt{21 \times(21-15) \times(21-13) \times(21-14)}$

$=84 \mathrm{~cm}^{2}$

BE is a perpendicular on AC

Now, area of triangle = 84 cm2

1/2 × BE × AC = 84 cm2

BE = 12 cm

The length of BE is 12 cm

 

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